Calculate $\int_0^\infty\frac{x^a}{x^3+1} dx$ using residues

Question

Although I've found similar integrals, I haven't found this.

Determine the values $a \geq 0 $ such that the integral:

$\int_0^\infty\frac{x^a}{x^3+1}dx $ is convergent.

For these values calculate it.

I have managed to check that $ a \in [0, 2) $, checking the interval $ (0, 1] $, in which there is no issues, because the limit tends to zero.

Also in $ [1, \infty) $, I get $$ \int_1^\infty x^{(a-3)} dx $$ and get that $ a < 2 $ or else it diverges because of the p-test.

Now to calculate by residue I'm unsure how to proceed, since it does not state that $ a $ is natural.

Searching with Approach0 or SearchOnMath returns some posts which contain this integral, such as: Computing an Integral using Contour Integration $\int_0^{\infty} \frac{x^{\alpha}}{x^3 + 1} dx$. — Martin Sleziak, Feb 25 '20 at 13:20
There is also a questions about a more general problem: Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$. — Martin Sleziak, Feb 25 '20 at 13:20
In case it helps, here is a proposed FAQ for searching: How to search on this site? — Martin Sleziak, Feb 25 '20 at 13:57

score 2 · Accepted Answer · answered Feb 25 '20 at 13:44

Calculate the integral $$\oint \frac{z^a}{z^3+1} \, dz = \oint \frac{\exp[a \log z]}{z^3+1} \, dz$$ over a suitable contour.

The contour could be a wedge from the origin to the real point $R$, then an arc of a circle to the point $\beta$ and back to the origin.

The point $\beta$ could be $R\exp[\frac{2 \pi i}{3}]$ containing one pole inside. Other similarly constructed contours will also work.

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